Answer :
Given:
Area of the virtual image of each square, A = 6.25 mm2
Area of each square, A0 = 1 mm2
Hence, the linear magnification of the object can be calculated as:
…(1)
Where, m = magnification
A = Area of virtual image
A0 = Area of each square
Putting values in equation (1) we get,
m =
⇒ m = 2.5
Also,
m = Image distance(v)/Object distance(u)
m = v/u
Therefore, v = m × u
= 2.5u …(2)
The magnifying glass has a focal length of, f = 10 cm
Applying the lens formula for lens we have:
…(3)
Where, f0 = focal length of the objective lens
v0 = Distance of image formation
u0 = Distance of object from objective
Putting values in the equation (3) we get,
⇒
⇒ u = -(1.5 × 10) / 2.5
Therefore, u = -6 cm
And v = 2.5u
⇒ v = 2.5 × 6
⇒ v = -15 cm
The virtual image cannot be seen by the eyes distinctly, because the image is formed at a distance of 15 cm, which is less than the near point (i.e. 25 cm) of a normal eye.
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